Skip to content
Choose the Best Answer · Q21

Q.A benzene ring bearing a -CH2-CH=CH2 (allyl) side chain →HCl\xrightarrow{HCl} (A) is

(a) the same allyl side chain retained (double bond intact) with a -Cl newly substituted on the ring at the position para to the side chain
(b) the same allyl side chain retained (double bond intact) with a -Cl newly substituted on the ring at the position ortho to the side chain
(c) both
(a) and
(b)
(d) the ring unchanged, with -Cl added across the side-chain double bond to give -CH2-CH(Cl)-CH3
Puducherry TnboardTextbookSubjectiveImportance★★★★★est
25% · 21/83 Questions
🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

Step 1. The substrate is allylbenzene, C6H5-CH2-CH=CH2 -- an aromatic ring carrying a three-carbon allyl side chain with an ISOLATED (non-conjugated) terminal double bond, separated from the ring by one -CH2- spacer.

Step 2. Options (a) and (b) both show the side chain's double bond left completely INTACT while a new -Cl appears directly on the aromatic RING (at the para or ortho position) -- that would require an electrophilic AROMATIC substitution, which needs a Lewis-acid catalyst (FeCl3, AlCl3, etc.) to generate a Cl+-type electrophile; plain HCl alone cannot do this to an unactivated benzene ring under ordinary conditions.

Step 3. What plain HCl CAN do readily, with no catalyst needed, is add across an isolated, ordinary alkene double bond by simple electrophilic (Markovnikov) addition -- exactly the side-chain -CH=CH2 unit here. …

Unlock everything free for 14 days

  • Full step-by-step solutions
  • Concept-first explanations
  • Methods, shortcuts & mistakes
  • PYQ mapping + timed mock tests

Full access for 14 days. No credit card required.