Answer :
Alright, let's break down the expected products for the [tex]\( \text{S}_\text{N}2 \)[/tex] reaction of 1-bromobutane with each of the following reagents:
### Part (a): [tex]\( \text{NaI} \)[/tex] (Sodium Iodide)
1-bromobutane reacts with sodium iodide in an [tex]\( \text{S}_\text{N}2 \)[/tex] reaction, where iodide ion ([tex]\( \text{I}^- \)[/tex]) acts as the nucleophile. In an [tex]\( \text{S}_\text{N}2 \)[/tex] reaction, the nucleophile attacks the carbon attached to the leaving group (bromine in this case) from the opposite side, displacing the bromide ion ([tex]\( \text{Br}^- \)[/tex]).
Product:
[tex]\[ \text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{I} \][/tex]
(butane to 1-iodobutane)
### Part (b): [tex]\( \text{KOH} \)[/tex] (Potassium Hydroxide)
When 1-bromobutane reacts with potassium hydroxide in an [tex]\( \text{S}_\text{N}2 \)[/tex] reaction, the hydroxide ion ([tex]\( \text{OH}^- \)[/tex]) acts as the nucleophile. The hydroxide attacks the carbon attached to the bromine, displacing the bromide ion.
Product:
[tex]\[ \text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{OH} \][/tex]
(butane to 1-butanol)
### Part (c): [tex]\( \text{NH}_3 \)[/tex] (Ammonia)
In this reaction, ammonia ([tex]\( \text{NH}_3 \)[/tex]) acts as the nucleophile. The nitrogen in ammonia attacks the carbon attached to the bromine in an [tex]\( \text{S}_\text{N}2 \)[/tex] reaction, displacing the bromide ion. The initial product is substituted butane attached with an ammonium group, which might further lose a proton to form a primary amine.
Product:
[tex]\[ \text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{NH}_3^+ \][/tex]
This often follows with:
[tex]\[ \text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{NH}_2 \][/tex]
(butane to 1-butylamine)
### Part (d): [tex]\( \text{H - C} \equiv \text{C} - \text{Li} \)[/tex] (Ethynyl Lithium)
In this case, the ethynyl anion [tex]\( \left(\text{HC} \equiv \text{C}^-\right) \)[/tex] is the nucleophile. It will attack the carbon attached to the bromine in an [tex]\( \text{S}_\text{N}2 \)[/tex] reaction, displacing the bromide ion.
Product:
[tex]\[ \text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{C} \equiv \text{CH} \][/tex]
(butane to 1-pentyne)
In summary, the expected products from the [tex]\( \text{S}_\text{N}2 \)[/tex] reactions of 1-bromobutane with the given reagents are:
(a) Sodium iodide [tex]\( \left(\text{NaI}\right) \)[/tex]: 1-iodobutane [tex]\( \left(\text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{I}\right) \)[/tex]
(b) Potassium hydroxide [tex]\( \left(\text{KOH}\right) \)[/tex]: 1-butanol [tex]\( \left(\text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{OH}\right) \)[/tex]
(c) Ammonia [tex]\( \left(\text{NH}_3\right) \)[/tex]: 1-butylamine [tex]\( \left(\text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{NH}_2\right) \)[/tex]
(d) Ethynyl lithium [tex]\( \left(\text{H - C} \equiv \text{C} - \text{Li}\right) \)[/tex]: 1-pentyne [tex]\( \left(\text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{C} \equiv \text{CH}\right) \)[/tex]
### Part (a): [tex]\( \text{NaI} \)[/tex] (Sodium Iodide)
1-bromobutane reacts with sodium iodide in an [tex]\( \text{S}_\text{N}2 \)[/tex] reaction, where iodide ion ([tex]\( \text{I}^- \)[/tex]) acts as the nucleophile. In an [tex]\( \text{S}_\text{N}2 \)[/tex] reaction, the nucleophile attacks the carbon attached to the leaving group (bromine in this case) from the opposite side, displacing the bromide ion ([tex]\( \text{Br}^- \)[/tex]).
Product:
[tex]\[ \text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{I} \][/tex]
(butane to 1-iodobutane)
### Part (b): [tex]\( \text{KOH} \)[/tex] (Potassium Hydroxide)
When 1-bromobutane reacts with potassium hydroxide in an [tex]\( \text{S}_\text{N}2 \)[/tex] reaction, the hydroxide ion ([tex]\( \text{OH}^- \)[/tex]) acts as the nucleophile. The hydroxide attacks the carbon attached to the bromine, displacing the bromide ion.
Product:
[tex]\[ \text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{OH} \][/tex]
(butane to 1-butanol)
### Part (c): [tex]\( \text{NH}_3 \)[/tex] (Ammonia)
In this reaction, ammonia ([tex]\( \text{NH}_3 \)[/tex]) acts as the nucleophile. The nitrogen in ammonia attacks the carbon attached to the bromine in an [tex]\( \text{S}_\text{N}2 \)[/tex] reaction, displacing the bromide ion. The initial product is substituted butane attached with an ammonium group, which might further lose a proton to form a primary amine.
Product:
[tex]\[ \text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{NH}_3^+ \][/tex]
This often follows with:
[tex]\[ \text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{NH}_2 \][/tex]
(butane to 1-butylamine)
### Part (d): [tex]\( \text{H - C} \equiv \text{C} - \text{Li} \)[/tex] (Ethynyl Lithium)
In this case, the ethynyl anion [tex]\( \left(\text{HC} \equiv \text{C}^-\right) \)[/tex] is the nucleophile. It will attack the carbon attached to the bromine in an [tex]\( \text{S}_\text{N}2 \)[/tex] reaction, displacing the bromide ion.
Product:
[tex]\[ \text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{C} \equiv \text{CH} \][/tex]
(butane to 1-pentyne)
In summary, the expected products from the [tex]\( \text{S}_\text{N}2 \)[/tex] reactions of 1-bromobutane with the given reagents are:
(a) Sodium iodide [tex]\( \left(\text{NaI}\right) \)[/tex]: 1-iodobutane [tex]\( \left(\text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{I}\right) \)[/tex]
(b) Potassium hydroxide [tex]\( \left(\text{KOH}\right) \)[/tex]: 1-butanol [tex]\( \left(\text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{OH}\right) \)[/tex]
(c) Ammonia [tex]\( \left(\text{NH}_3\right) \)[/tex]: 1-butylamine [tex]\( \left(\text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{NH}_2\right) \)[/tex]
(d) Ethynyl lithium [tex]\( \left(\text{H - C} \equiv \text{C} - \text{Li}\right) \)[/tex]: 1-pentyne [tex]\( \left(\text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{C} \equiv \text{CH}\right) \)[/tex]
