Answer
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Hint: Organic compounds are the compounds which in solid, liquid, and gaseous state contain carbon in its molecule. A large number of organic compounds are there and a systematic classification is required. Organic compounds are classified as open-chain (acyclic), closed chain (cyclic). Open chain is classified into two types: straight-chain and branched-chain. The closed chain is classified into two types: Homocyclic compounds and Heterocyclic compounds. Homocyclic compounds are classified into two types: Alicyclic compounds and Aromatic compounds.
Complete step by step answer:
When 3, 3- dimethyl-1-butene reacts with HCl it gives 2-chloro-2, 3-dimethyl butane as the major product. The reaction along with the mechanism is written below:
Mechanism:
Step 1: In this step, the double bond in 3, 3- dimethyl-1-butene breaks and it forms a secondary carbocation, and the ${H^ + }$ ion of HCl gets added to the negative carbon.
Step 2: In this step one methyl group gets shifted to secondary carbocation to form tertiary carbocation which is the most stable carbocation.
Step 3: The $C{l^ - }$ ion will be added to the tertiary carbocation to form the major product 2-chloro-2, 3-dimethyl butane.
Therefore the major product is 2-chloro-2, 3-dimethyl butane.
Hence, the correct option is B.
Note: When a group X, having electronegativity higher than carbon is attached to it, $\left( {C - X} \right)$ bond breaks heterolytically. The shared pair of electrons is given to the atom other than the carbon. Thus, the carbon atom acquires a positive charge. It has only six electrons in its valence shell. Such a type of ion of carbon is known as carbocation or carbonium ion. The positively charged ion of the carbocation is $s{p^2}$ hybridized.
Complete step by step answer:
When 3, 3- dimethyl-1-butene reacts with HCl it gives 2-chloro-2, 3-dimethyl butane as the major product. The reaction along with the mechanism is written below:
Mechanism:
Step 1: In this step, the double bond in 3, 3- dimethyl-1-butene breaks and it forms a secondary carbocation, and the ${H^ + }$ ion of HCl gets added to the negative carbon.
Step 2: In this step one methyl group gets shifted to secondary carbocation to form tertiary carbocation which is the most stable carbocation.
Step 3: The $C{l^ - }$ ion will be added to the tertiary carbocation to form the major product 2-chloro-2, 3-dimethyl butane.
Therefore the major product is 2-chloro-2, 3-dimethyl butane.
Hence, the correct option is B.
Note: When a group X, having electronegativity higher than carbon is attached to it, $\left( {C - X} \right)$ bond breaks heterolytically. The shared pair of electrons is given to the atom other than the carbon. Thus, the carbon atom acquires a positive charge. It has only six electrons in its valence shell. Such a type of ion of carbon is known as carbocation or carbonium ion. The positively charged ion of the carbocation is $s{p^2}$ hybridized.
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