Home
Class 12
CHEMISTRY
The rate of formation of alcohols from C...

The rate of formation of alcohols from `CH_(3)Br, CH_(3)CH_(2)Br, (CH_(3))_(2) CHBr` which follow `SN_2` mechanism is :

A

`CH_3Br `gt` CH_3CH_2Br `gt` (CH_3)_2CHBr`

B

`CH_3Br `gt` (CH_3)_2CHBr `gt` CH_3CH_2Br`

C

`(CH_3)_2CHBr `gt` CH_3CH_2Br `gt` CH_3Br`

D

`(CH_3)_2CHBr `gt` CH_3Br `gt` CH_3CH+_2Br`

Text Solution

Verified by Experts

The correct Answer is:
A
The steric hindrance causes the decrease in the rate of `S_N2` reaction.
Promotional Banner

Similar Questions

Explore conceptually related problems

Consider the reaction, CH_(3)CH_(2) CH_(2)Br + NaCN to CH_(3)CH_(2)CH_(2)CN + NaBr . This reaction will be the faster in:

Arrange the following compounds in the order of reactivity with alcoholic silver nitrate, C_(6)H_(5)CH_(2)CH_(2)Br (I), C_(6)H_(5)CHBrCH_(3)(II) and C_(6)H_(5)CH = CHBr (III)

Give the IUPAC names of the following compounds: CH_3CH(CI)CH(Br)CH_3

For the following reactions: (A) CH_3 CH_2 CH_2 Br + KOH to CH_3 CH = CH_2 + KBr + H_2O Which of the following statement is correct?

Position of bromine in the compound in CH_(3)CH = CHC (Br)(CH_(3))_(2) can be classified as………….

Which are the branches? CH_3 CH_2 CH(CH_2 CH_2 CH_3) CH_2 CH(CH_3) CH_2 CH_3

Write the IUPAC name of the following compounds? i) CH_(3)CH=CH_(2) ii) CH_(3)CH_(2)CH=CHCH_(3)