Question

Que 1: Compound A, CH, Cl, gives compound B, C,Ho, when treated with AICI3. The 'H- NMR spectrum of B is given below. Show the structure of compounds A and B. Also show the mechanism for the reaction. let Hunt 10 Ppm

this is quite tough question please help me in this que as I have no idea even how should I start:(

Answer 1

Degree of Unsaturation (DOU) is also known as Double Bond Equivalent. If the molecular formula is given, DoU can be calculated as follows

DoU   =   2C+2+N−X−H

               2

C is the number of carbons

N is the number of nitrogens

X is the number of halogens (F, Cl, Br, I)

H is the number of hydrogens

Compound A has molecular formula of C9H11Cl

Degree of unsaturation in the molecule = 4

While compound B has molecular formula of C9H10

Degree of unsaturation = 5

It indicates that compound possess either double bond/cyclic structure/carbonyl functionality.

Since oxygen is absent, carbonyl functionality is not present. While compound B does not possess alkene proton as per 1H NMR. Thus compound have a cyclic structure which constitutes one site of unsaturation

Compound B NMR data as follows

1H NMR data:

NMR peak value (ppm)	Protons	Indication
7.1-7.2	4H multiplet	Four aromatic protons. Pattern indicates presence of ortho substitution
3.0	4H, triplet	Indicates two equivalent CH2 attached to CH2 and quaternary carbon
2.0	2H, quintet	CH2 attached to two CH2

Data indicates presence of –CH2-CH2-CH2- linkage in the molecule

Chemical Formula: CH11C1 2,3-dihydro-1H-indene Chemical Formula: CH 10 The reaction is Friedel-Craft alkylation reaction.