Answer : The correct answer for a) 4-bromo-2-iodo-4-methyl pentane and b)5-bromo-2-ethoxy-2-methyl pentane.
A) Reaction with NaI :
Reaction of alkyl halide with NaI is known as Finkelstein Reaction . The acetone is used as solvent . It involves bimolecular nucleophillic substitution rmechanism (SN²) . There is replecement of one halogen with other occurs .
The incoming Nucleophile(Nu⁻) (halide) attacks on carbon from back side , while the leaving group (halide) leaves the compound from front side , simultaneously. The product so formed have is inverted .(Image)
NaI releases I⁻ ion which act as nucelophile and attacks on C1 carbon and Br⁻ from C1 carbon is released . Out of two bromines at C1 and C4 carbons , C1 is primary carbon which is less sterically hindered while C-4 is tertiary carbon and sterically hindered . So it is easy for incoming Nu⁻ to attack on C1 carbon .So Br⁻ is repleaced by I⁻.
1,4-dibromo-4-methylpentane + NaI → 4-bromo-1-iodo-4-methylpentane
The product formed from reaction between 1,4-dibromo-4-methylpentane and NaI is 4-bromo-1-iodo-4-methylpentane . (Image)
B) Reaction with AgNO3 :
Reaction of alkyl halide with AgNO3 in ethanol takes place via SN¹ ( unimolecular nucleophilic substitution ) mechanism . In this leaving group(halide) leaves from alkyl halide forming an intermediate carbocation species . The incoming Nu⁻ attack on this carbocation.
AgNO3 reacts releases Ag⁺ion which abstract Br⁻ of C-4 carbon from 1,4-dibromo-4-methylpentane. THis forms tertiary carbocation which is more stable than carbocation formed by removal of Br from C-1 . The ethanol being more Nucleophilic than NO₃⁻ (from AgNO₃), attacks on this carbocation .(Image )
The product formed as a result is 5-bromo-2-ethoxy-2-methyl pentane.
Answer:
3NaOH (aq) + Fe(NO₃)₃ (aq) → Fe(OH)₃ (s) + 3NaNO₃ (aq)
Explanation:
Step 1: RxN
NaOH (aq) + Fe(NO₃)₃ (aq) → Fe(OH)₃ (s) + NaNO₃ (aq)
Step 2: Balance RxN
We need 3 OH's on both sides.
We also need 3 NO₃'s on both side.
- This will make it so we also need 3 Na's on both side
3NaOH (aq) + Fe(NO₃)₃ (aq) → Fe(OH)₃ (s) + 3NaNO₃ (aq)
Answer:
Explanation:1. NaNH2 (1-Butene)
CH3CH2CH2CH2Cl --------------> CH3CH2CH=CH2 + HCl (elimination reaction)
2. Br2, CCl4
CH3CH2CH=CH2 ---------------> CH3CH2CH(Br)CH2Br (Simple addition Reaction)
3. NaNH2 (1-Butyne)
CH3CH2CH(Br)CH2Br ----------------> CH3CH2C≡CH + 2HBr
Sodamide (NaNH2) is a very strong base and generally results in Terminal Alkynes when treated with Vicinal Dihalides.
Alcoholic KOH on the other hand results in the formation of Alkynes with triple bonds in the middle of the molecule.