Common Organic Questions

Question 1

why are alkanes unreactive?

Answer 1

• non-polar, does not attract nucleophiles and electrophiles
• C−C and C−H bonds very strong, ↑ e neede to overcome

nucleophiles & electrophiles are polar

Question 2

boiling pt for alkanes

Answer 2

• structure: simple molecular structure non polar
• imf: id-id
• ↑length of chain: e- cloud larger & polarisable → more e needed to overcome stronger id-id →↑bp
• branching: smaller surface area of contact→ ↓e required to overcome the weaker & less extensive id-id b/w molecules

Question 3

Explain how the cis-trans isomerism arises.

alkene topic

Answer 3

• two different groups of atoms are bonded to each of the C atoms involved in the C=C
• restricted rotation of C=C due to the presence of pi bonds

Question 4

reactivity of arenes to electrophilic attack

arenes are cpds containing benzene ring

Answer 4

• has activating group: ↑ electron density of the ring → ↑ susceptible to electrophilic attacks
• has deactivating group: ↓ electron density of the ring → ↓ susceptible to electrophilic attacks

Question 5

Why benzene does not undergo addition reactions

Answer 5

• addition reactions will cause the disruption of delocalised pi electrons cloud
• causing benzene to lose the resonance stability

Question 6

boiling pt for halogenoalkanes

Answer 6

• structure: simple molecular, polar
• imf: id-id & pd-pd
• ↑length of chain: e- cloud larger & polarisable → more e needed to overcome id-id
• comparison b/w different halogens pls elab

Question 7

reactivity of halogeno alkane towards nucleophile

Answer 7

• bond strength of the C – X ↓ from Cl to Br to I
• effectiveness of the orbital overlap between the C atom and halogen atom decreases as
• size of orbital increases from Cl to I
• nucelophilic substitution increases from Cl to I

Question 8

why halogeno arenes dont undergo nucleophilic sub?

halogeno arenes are benzene ring with halogen attached

Answer 8

• delocalisation of lone pair of e- on halogen atom into the benzene ring
• resulted carbon–halogen bond having partial double bond character
• stronger and more difficult to break
• rear side of the carbon–halogen bond blocked by benzene ring
• pi electron cloud of benzene ring repels the lone pair of e- of incoming nucleophile

Question 9

why halogeno arenes dont undergo addition rxn?

Answer 9

resonance stability of the benzene ring needs preserved

Question 10

what favours Sn1 rxn?

halogeno derivative topic

Answer 10

• Favoured by stability of carbocation formed
  (more steric hindrance)
• rate of formation of carbocation:
• ↑ e- donating grp disperse +ve charge of carbonation
• carbocation stabilised and formed faster
• tertiary bromoalkane with greater steric hindrance around the C bonded to Br, making SN 2 mechanism less favourable.
• strength of carbon-halogen bond, hinders the approach of the nucleophile
• down grp 17, size of atom increase, valence orbital becomes larger and more diffused, less effective overlap, less E needed to break the weaker carbon halogen bond​

Question 11

what favours Sn2 rxn?

halogeno derivative topic

Answer 11

• Favoured by less steric hindrace,
• steric effect of R group → carbonyl carbon less **electron deficient **→ nucleophile is less attracted
• backside attack of nucleophile also hindered by bulky R grp
• strength of carbon-halogen bond:
• down grp 17, size of atom increase, valence orbital becomes bigger and more diffused, less effective overlap, less E needed to break the weaker carbon halogen bond​

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Question 12

why is there racemic mixture in Sn1 rxn?

halogeno derivative topic

Answer 12

• carbocation arranged in trigonal planar
• allows of nucleophilic attack from both side of the plane with equal probabilities
• equal amt of 2 enantiomers of product formed
• two enantiomers rotate plane polarised light to an equal angle in opposite direction, → rotation is cancelled out

Question 13

boiling pt for hydroxyl cpd

Answer 13

• structure: simple molecular, polar
• imf: id-id, pd-pd, H2 bond
• intermolecular hydrogen bond (↑bp)
• no. of OH grps: extensiveness of the intermolecular hydrogen bonding ↑, ↑ e needed to overcome stronger h2 bond
• have intramolecular hydrogen bonds: limits the number of sites available for intermolecular hydrogen bonding (↓bp)
• ↑length of chain: e- cloud larger & polarisable → more e needed to overcome id-id
• linear: more extensive id-id between its molecules

Question 14

why phenol dont undergo rxn with nucleophilic sub?

Answer 14

• delocalisation of lone pair of e- on oxygen atom into the benzene ring
• resulted C-O bond having partial double bond character
• stronger and more difficult to break
• rear side of the carbon–oxygen bond blocked by benzene ring
• pi electron cloud of benzene ring repels the lone pair of e- of incoming nucleophile

Question 15

acidity of phenol vs alcohol

IMPORTANT

Answer 15

alcohol weaker:
* electron-donating alkyl grp
* intensify -ve charge on O atom of alkoxide ion
* alkoxide destabilised, alchol dissociates to a smaller extent, producing less H+

phenol stronger:
* p-orbital of oxygen overlaps with the pi-electron cloud of the benzene ring
* -ve charge on the oxygen atom of the phenoxide ion delocalised into benzene ring
* phenoxide stabilised as charge is dispersed
* phenol dissociates to a greater extent, producing more H+

more stable to congugate base, stronger the acid

Question 16

why aldehyde undergo nucleophilic sub faster than ketone?

primary alcohol vs secondary alcohol

Answer 16

• aldehyde has 1 e- donating grp, ketone has 2 e- donating grp
• partial positive charge on the carbonyl carbon in ketone dispersed to greater extent
• ketone less electron deficient and less susceptible to nucleophilic attack
• (or) additional bulky -CH3 grp ↑ steric hindrance about the carbonyl carbon in ketone
• hinders the approach of the attacking nucleophile

Question 17

boiling point for carboxylic acid

Answer 17

• structure: simple molecular, polar
• imf: id-id, pd-pd, H2 bond
• intermolecular hydrogen bond (↑bp) (compare with alcohols)
• H2 bond stonger in carbo acid as OH grp is more polarised due to e- withdrawing C=O grp
• ↑length of chain: e- cloud larger & polarisable → more e needed to overcome id-id
• linear: more extensive id-id between its molecules

Question 18

boiling point for ionic cpd

Answer 18

• structure: giant ionic lettice structure
• imf: strong ionic bonds
• strong electrostatic attraction b/w X+ & Y-
• requires the most energy to break

Question 19

formation of H2 bond

Answer 19

1. H atom bonded directly to N, O or F in one molecule
2. N, O or F with a lone pair of electrons in another molecule.

N,O,F are highly electronegative atoms

O have partial -ve charge, H have partial +ve charge, draw H2 bond connecting the lone pair of e- on O to H atom

Question 20

why solubility decreases when length of carbo acid increases?

Answer 20

1. state interactions:
   * id-id b/w R grps of acid and H2 bond b/w COOH grp (solute-solute)
   * H2 bond b/w water molecules (solvent-solvent)
   * H2 bond b/w COOH grp and water molecules (solute-solvent)
2. As length↑, id-id b/w R grps of acid more significant, interferes with H2 bond b/w water molecules
3. interactions b/w water and acid becomes more predominantly id-id (initially it was more predominantly h2 bond with water)
4. e released when forming id-id with water insufficient to overcome H2 bond with water and more predominant id-id b/w acid molecules

Question 21

acidity of OH containing cpd

alcohol<phenol<carbo acid<clCOOH

Answer 21

alcohol:
* e- donating alkyl grp intensify -ve charge on O atom in alkoxide ion
* alkoxide ion least stable, alcohol least acidic
phenol:
* p-orbital O overlap with pi e- cloud of benzene, -ve charge of O delocalise into ring
* resonance stabilisation not as great as in carboxylate ion
* -ve charge dispersed, phenoxide ion stabilised
carbo acid:
* has 2 equivalent resonance structures ( -O and =O)
* -ve charge delocalised over 2 highly electronegative O atoms
* -ve charged more effectively dispersed
halogen carbo acid:
* electronegative halogen exert (greater) e- withdrawing effect, dispersing -ve charge on O
* carboxylate ion most stabilised

conguate base more stable, acid more acidic

Question 22

ease of hydrolysis of chloro-compounds

B-COCl vs B-CH2Cl vs B-Cl

B represents bezene

Answer 22

dependent on:
* e- deficiency of C bonded to halogen
* steric factors that hinder attack by nucleophile
COCl:
* carbonyl carbon highly e- deficient as bonded to 2 e-ve atoms (O&Cl)
* carbon is sp2 hybridised (triginal planar), less steric hindrance
* most susceptible to nucleophilic attack
CH2Cl:
* carbonyl carbon less e- deficient as only bonded to 1 e-ve atom (Cl)
* sp3 hybirdised (tetrahedral) more steric hindrace
* less susceptible to N attack
Cl:
* p orbital of Cl overlap with pi e- cloud of B ring, lone pair e- from p orbital delocalise into ring
* C-Cl bond partial double bond character
* benzene ring also repel attcking nucleophile

attack by :OH nucleophile to convert acyl chloride to carboxylic acid

Question 23

compare boiling pt for alkane, COOH, amine, amino acid

Answer 23

Alkane:
* non polar, least e needed to overcome weak id-id, lowest mp

COOH and Amine:
* both polar, have stronger id-id, pd-pd and H2 bond
* O more e-ve than N, larger dipole moment, O-H bond more polar than N-H bond, H2 bond in COOH stronger
* COOH higher bp than Amine

Amino Acid:
* structure: zwitterions held tgt by strong ionic bonds
* largest amt of e to overcome, highest bp

Question 24

Why are amides neutral?

SIGMA IMPTTTT

Answer 24

lone pair of e- on N in –NHCO not available for coordination to H+ as it is delocalised into the e- withdrawing C=O grp

nothing to do with amides being zwitterions

Question 25

basicities of amines

Answer 25

dependent on: availability of lone pair of e- on N to form a dative covalent bond with a proton e- donating grp: * make lone pair of e- **more available** for donation to H+ ion * **p orbital** of N overlap with pi e- cloud of B ring, lone pair e- from p orbital **delocalise** into ring, **less available** to form dative covalent bond with **proton** to **form an acidic group** Electronegative species attached to benzene ring further decreases the e- density of N atom, lone pair of e- least available for coordination with proton

Question 26

why is NaOH needed for condensation phenols?

Answer 26

To react with phenol to form pehnoxide ion which is a stronger nucleophile

Question 27

how does cis-trans isomerism arise?

Answer 27

presence of restricted rotation about the C=C double bond [1] and that each C in the C=C is bonded to 2 different groups of atoms [1].

Question 28

Why is phenol more susceptible to electrophili substitution?

Answer 28

The −OH group directly bonded to the benzene is a **strongly activating group** as lone pair of electrons on the O atom can interact with the delocalised π-electron cloud of the benzene ring and delocalises into the ring. This **increases the electron density** in the benzene ring and makes it **more susceptible to electrophiles**.

Question 29

Why phenol does not react with carboxylic acid?

Answer 29

* overlap b/w O atom and pi e- cloud of benzene ring * lone pair of e- from O delocalised into the benzene ring * reducing the availability of lone pair of e- to attack a carboxylic acid

Question 30

Why phenol does not react with carboxylic acid?

Answer 30

* overlap b/w O atom and pi e- cloud of benzene ring * lone pair of e- from O delocalised into the benzene ring * reducing the availability of lone pair of e- to attack a carboxylic acid

Question 31

Explain why LiAlH 4cannot be used to react with C=C?

Answer 31

C=C is non-polar and is not susceptible to nucleophilic attack by negatively charged hydride ions from LiAlH4.