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

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

Explanation

Solution

According to the Saytzeff rule, in dehydrogenation reaction, the preferred product is that alkene which has the greater number of alkyl groups attached to the doubly bonded carbon atoms. This implies that base induced elimination (E2)\left( {{E_2}} \right) will lead predominantly.

Complete step by step answer:
In the given reaction we have to find out AA and BB which can be determine in multi steps
In step11: pro-1-ene (CH3−CH=CH2)\left( {C{H_3} - CH = C{H_2}} \right) react with hydrogen bromide (HBr)\left( {HBr} \right)it undergoes addition reaction and give 1-bromopropane (CH3−CH2−CH2−Br)\left( {C{H_3} - C{H_2} - C{H_2} - Br} \right) . The formation of 2-bromopropane is due to anti markovnikov rule (according to anti markovnikov rule bromide ion (Br∙)\left( {\mathop {Br}\limits^ \bullet } \right) goes to that carbon atom which has more number of hydrogen).
In step 2: 1-Bromopropane (CH3−CH2−CH2−Br)\left( {C{H_3} - C{H_2} - C{H_2} - Br} \right) reacts with alcoholic potassium hydroxide(KOH acl)\left( {KOH\,acl} \right) it undergoes dehydrohalogenation reaction and gives pro-1-ene (CH3−CH=CH2)\left( {C{H_3} - CH = C{H_2}} \right).In this reaction alcoholic potassium hydroxide (Alcoholic KOHKOH is more strong base, then aqueous. It is used for preparation alkenes from alkyl halides) act as a reducing agent, it reduces bromopropane to pro-1-ene. This is because removable hydrogen from β−\beta - carbon and halides from α−\alpha - carbon atoms in the presence of acl. KOHacl.\,KOHresult in the formation of alkenes. This reaction is known as β−\beta - elimination reaction.

Note:
β−\beta - Elimination (beta-elimination), a chemical reaction in which atoms or groups are lost from adjacent atoms resulting in formation of a new pi-bond.
Saytzeff’s rule, According to this rule although alkene synthesis leads to form more than one product, the more substituted alkene is the major product.
HBrHBr Breaks into free radicals because the electronegativity of hydrogen and bromine atoms are similar there the charge is partially separated on them in the presence of peroxide.