Answer :
To determine the major product of the reaction sequence involving the given reagents and conditions, let's go through the steps one by one:
1. Reaction with [tex]\( Br_2 \)[/tex] and heat:
When a compound, typically an alkane, is treated with bromine [tex]\((Br_2)\)[/tex] and heat, a free radical halogenation reaction occurs. This process involves the formation of a bromine radical from the homolytic cleavage of [tex]\( Br_2 \)[/tex] due to the thermal energy provided by heat. The bromine radical then abstracts a hydrogen atom from the alkane to form HBr and create an alkyl radical. The alkyl radical subsequently reacts with another [tex]\( Br_2 \)[/tex] molecule to form the alkyl bromide and regenerate a bromine radical.
The regioselectivity of the halogenation is such that the more stable radical intermediate is preferred. For instance, tertiary radicals are more stable than secondary, which in turn are more stable than primary radicals. Thus, the bromine is more likely to attach at a position that gives the most stable radical intermediate.
2. Reaction with 2 equivalents of NaCN in DMF:
Sodium cyanide (NaCN) in dimethylformamide (DMF) is a source of the cyanide ion [tex]\((CN^-)\)[/tex], which is a good nucleophile. In this step, the nucleophile [tex]\( CN^- \)[/tex] will perform an [tex]\( S_N2 \)[/tex] substitution reaction on the alkyl bromide formed in the first step.
Given that the second reaction step involves 2 equivalents of NaCN, it suggests that there are two bromine atoms introduced during the first step, resulting in a dibrominated product. Both bromines will be replaced by [tex]\((CN^-)\)[/tex] ions resulting in a dicyanide product.
Putting these steps together, the reaction sequence transforms an initial alkane (like cyclohexane, for example) first into a dibromoalkane due to the halogenation process, followed by substitution with cyanide ions to form a dicyanide product. Here is a step-by-step outline:
### Step 1: Halogenation with [tex]\( Br_2 \)[/tex], heat
1) Starting with an alkane (for simplicity, let's consider cyclohexane):
[tex]\[ \text{Cyclohexane} + Br_2 \xrightarrow{\text{heat}} \text{1,2-Dibromocyclohexane} \][/tex]
### Step 2: Substitution with NaCN in DMF
2) 1,2-Dibromocyclohexane undergoing nucleophilic substitution:
[tex]\[ \text{1,2-Dibromocyclohexane} + 2 \, \text{NaCN} \xrightarrow{\text{DMF}} \text{1,2-Dicyanocyclohexane} \][/tex]
Thus, the major product of the reaction sequence is 1,2-dicyanocyclohexane.
When considering stereochemistry, the [tex]\(Br_2\)[/tex] addition is initially a trans addition to the double bond if double bonds were involved due to anti-addition mechanisms; otherwise, for a simple radical mechanism as shown, 1,2-addition without specific cis/trans requirements applies for the intermediate. Finally, the [tex]\(S_N2\)[/tex] substitution mechanism with NaCN ensures inversion of configuration at each substitution site, which typically does not alter the trans nature if it was initial.
1. Reaction with [tex]\( Br_2 \)[/tex] and heat:
When a compound, typically an alkane, is treated with bromine [tex]\((Br_2)\)[/tex] and heat, a free radical halogenation reaction occurs. This process involves the formation of a bromine radical from the homolytic cleavage of [tex]\( Br_2 \)[/tex] due to the thermal energy provided by heat. The bromine radical then abstracts a hydrogen atom from the alkane to form HBr and create an alkyl radical. The alkyl radical subsequently reacts with another [tex]\( Br_2 \)[/tex] molecule to form the alkyl bromide and regenerate a bromine radical.
The regioselectivity of the halogenation is such that the more stable radical intermediate is preferred. For instance, tertiary radicals are more stable than secondary, which in turn are more stable than primary radicals. Thus, the bromine is more likely to attach at a position that gives the most stable radical intermediate.
2. Reaction with 2 equivalents of NaCN in DMF:
Sodium cyanide (NaCN) in dimethylformamide (DMF) is a source of the cyanide ion [tex]\((CN^-)\)[/tex], which is a good nucleophile. In this step, the nucleophile [tex]\( CN^- \)[/tex] will perform an [tex]\( S_N2 \)[/tex] substitution reaction on the alkyl bromide formed in the first step.
Given that the second reaction step involves 2 equivalents of NaCN, it suggests that there are two bromine atoms introduced during the first step, resulting in a dibrominated product. Both bromines will be replaced by [tex]\((CN^-)\)[/tex] ions resulting in a dicyanide product.
Putting these steps together, the reaction sequence transforms an initial alkane (like cyclohexane, for example) first into a dibromoalkane due to the halogenation process, followed by substitution with cyanide ions to form a dicyanide product. Here is a step-by-step outline:
### Step 1: Halogenation with [tex]\( Br_2 \)[/tex], heat
1) Starting with an alkane (for simplicity, let's consider cyclohexane):
[tex]\[ \text{Cyclohexane} + Br_2 \xrightarrow{\text{heat}} \text{1,2-Dibromocyclohexane} \][/tex]
### Step 2: Substitution with NaCN in DMF
2) 1,2-Dibromocyclohexane undergoing nucleophilic substitution:
[tex]\[ \text{1,2-Dibromocyclohexane} + 2 \, \text{NaCN} \xrightarrow{\text{DMF}} \text{1,2-Dicyanocyclohexane} \][/tex]
Thus, the major product of the reaction sequence is 1,2-dicyanocyclohexane.
When considering stereochemistry, the [tex]\(Br_2\)[/tex] addition is initially a trans addition to the double bond if double bonds were involved due to anti-addition mechanisms; otherwise, for a simple radical mechanism as shown, 1,2-addition without specific cis/trans requirements applies for the intermediate. Finally, the [tex]\(S_N2\)[/tex] substitution mechanism with NaCN ensures inversion of configuration at each substitution site, which typically does not alter the trans nature if it was initial.
