Answer : The correct option is, (B) 6 mole
Explanation :
Given moles of 
= 6 moles
Given moles of 
= 6 moles
First we have to calculate the limiting and excess reagent.
The balanced chemical reaction is,
From the given balanced reaction, we conclude that
As, 1 moles of react with 2 moles of
So, 6 moles of react with 
moles of
From this we conclude that, is an excess reagent and is a limiting reagent because the given moles are less than the required moles and it limits the formation of product.
Thus, the number of moles of NaOH used up in the reaction = Required moles of NaOH - Given moles of NaOH
The number of moles of NaOH used up in the reaction = 12 - 6 = 6 moles
Therefore, the number of moles of NaOH used up in the reaction will be, 60 moles
526 L O2 x 1 mol O2 / 22.4 L = 23.5 mol O2
Density * Volume = Mass
Now we substitute the values in.
19.3 g/cm^3 + 20 cm^3 = 386 g
Mass = 386 g
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.
