Please wait...
/
-
The yield of alkyl bromide obtained as a result of heating the dry silver salt of carboxylic acid with bromine in CCI4 is
The reaction follows free radical mechanism and alkyl free radical is formed in the propagation step as
Hence, stability of alkyl free radical (3° > 2° > 1°) determine the reactivity.
Which is incorrect about Hunsdiecker's reaction?
Except F2, almost all halogens react with RCOOAg giving alkyl halide via Hunsdiecker reaction.
With l2 if RCOOAg is in excess, R— I formed in the first step reacts further with unreacted salt to give ester asR—COOAg + R—I → R—COOR + AgI
The major product of the following reaction is
Free radical bromination occur. Preferably at highest degree carbon where most stable free radical is formed.
Racemic mixture is obtained due to the halogenation of
If free radical halogenation generate a chiral carbon, racemic mixture of halides are always formed due to equal probability of halogenation from both sides of planar free radical.
The reaction of SOCI2 on alkanols to form alkyl chlorides gives good yields because
The gaseous byproducts escape out on its own continuously driving the reaction in forward direction.
Addition of bromine on propene in the presence of brine yields a mixture of
Nucleophilic attack in step-ll occur at the carbon atom which can better accommodate the positive charge. Hence, attack of Br- or Cl- in second step occur at 2° carbon rather that at 1° carbon.
Correct (-)
Wrong (-)
Skipped (-)