Scoring Scheme: 3-3-2-1 When you use the half reaction method to balance the redox reaction between one mole of NBS and ascorbic
acid to give L-dehydroascorbic acid and succinimide, how many moles of electrons are transferred?
1 answer:
<span>In the redox reaction between one mole of NBS and ascorbic acid, ascorbic acid undergoes oxidation reaction to form </span><span>L-dehydroascorbic acid.
On other hand, N-bromo</span>succinimide (NBS) undergoes reduction to form <span>succinimide.
Electron transferred during this process is two. Complete reaction is attached below for quick reference.
Thus, one molecule of NBS reacts with one molecule of </span>ascorbic acid, during this redox reaction. For each molecule reacting, two electrons are transferred.
Thus, when one mole of NBS reacts with one mole of ascorbic acid, 2 x Na moles of electrons will be transferred. It may be noted here that, Na = Avagadro's number = 6.022 X 10^23.
On other hand, N-bromo</span>succinimide (NBS) undergoes reduction to form <span>succinimide.
Electron transferred during this process is two. Complete reaction is attached below for quick reference.
Thus, one molecule of NBS reacts with one molecule of </span>ascorbic acid, during this redox reaction. For each molecule reacting, two electrons are transferred.
Thus, when one mole of NBS reacts with one mole of ascorbic acid, 2 x Na moles of electrons will be transferred. It may be noted here that, Na = Avagadro's number = 6.022 X 10^23.
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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.
