Solveeit Logo
Login

Question

Question: Identify the A and B in the following reaction: \(C{H_3} - CH = C{H_2}\xrightarrow{{HBr}}A\xrighta...

Identify the A and B in the following reaction:
CH3−CH=CH2→HBrA→KOH aclBC{H_3} - CH = C{H_2}\xrightarrow{{HBr}}A\xrightarrow{{KOH{\text{ }}acl}}B

Explanation

Solution

Markovnikov rule states that in addition of protic acid (HX)\left( {HX} \right) to an unsymmetrical alkene, acidic hydrogen gets attached to that carbon atom of double bond which has more hydrogen atoms and halide group gets attached to that carbon atom of double bond which is attached with more alkene groups.

Complete step by step answer:
Reaction given in this question is CH3−CH=CH2→HBrA→KOH aclBC{H_3} - CH = C{H_2}\xrightarrow{{HBr}}A\xrightarrow{{KOH{\text{ }}acl}}B and we have to identify A and B in this reaction.
In the first part of the reaction, that is, CH3−CH=CH2→HBrAC{H_3} - CH = C{H_2}\xrightarrow{{HBr}}A anti markovnikov rule will be followed. Markonikov rule states that in addition of protic acid (HX)\left( {HX} \right) to an unsymmetrical alkene, acidic hydrogen gets attached to that carbon atom of double bond which has more hydrogen atoms and halide group (X)\left( X \right) gets attached to that carbon atom of double bond which is attached with more alkene groups. In this reactant, the carbon atom of CH2C{H_2} has more hydrogen atoms as compared to CHCH. So, according to anti markovnikov rule hydrogen atom will be attached to CH2C{H_2} group and bromine atom will be attached to CHCH group. So, the resulting compound that is A will be:

The chemical formula of this compound is C3H7Br{C_3}{H_7}Br.
So the first step of reaction is:
CH3−CH=CH2→HBrCH3−CHBr−CH3C{H_3} - CH = C{H_2}\xrightarrow{{HBr}}C{H_3} - CHBr - C{H_3}
Second step of reaction is:
CH3−CHBr−CH3→KOH aclBC{H_3} - CHBr - C{H_3}\xrightarrow{{KOH{\text{ }}acl}}B
In this reaction halogen of the reactant will be replaced with OHOH group of KOHKOH and KBrKBr will be released. So, B is C3H7OH{C_3}{H_7}OH and the complete reaction is:
CH3−CH=CH2→HBrCH3−CHBr−CH3→KOH aclCH3−CHOH−CH3C{H_3} - CH = C{H_2}\xrightarrow{{HBr}}C{H_3} - CHBr - C{H_3}\xrightarrow{{KOH{\text{ }}acl}}C{H_3} - CHOH - C{H_3}
Hence A and B are C3H7Br{C_3}{H_7}Br and C3H7OH{C_3}{H_7}OH respectively.

Note:
Anti-markovnikov rule is opposite to the markovnikov rule. According to Anti-markovnikov rule hydrogen atom of HXHX gets attached to that carbon atom of double bonded carbon which has less number of hydrogen atoms and halogen gets attached to that carbon atom of double bonded carbon atom which has more number of hydrogen atoms.