Make sure the equation is always balanced first. (It is balanced for this question already) 6.022 x 10^23 is Avogadro’s number. In one mole of anything there is always 6.022 x 10^23 molecules, formula units, atoms. For one mol of an element/ compound use molar mass (grams).
Multiply everything on the top = 8.61x10^47
Multiple everything on bottom= 1.20x10^24
Divide top and bottom = 7.15x10^23
Answer: 7.15x10^23 mol SO2
<span> mass of glucose = 0.055 *165 = 9.075 g
vol of methyl alc = 0.185 * 1.87 = 0.346 L = 346 ml
% NaCl ( m/v ) = mass NaCl * 100/ vol of soln
or Vol of Soln = mass NaCl / % NaCl (m/v)
= 32.1 * 100 / 6 = 535 ml the total vol of soln</span>
Answer:
The reaction in which heat is absorbed from the surrounding is called endothermic reaction.
so
I think it's answer is 2nd option.
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.