carboxylic acids

Question 1

small Ka

Answer 1

weaker acid

Question 2

put these in order :
CA
ethanol
phenol
water

Answer 2

CA > phenol > water > ethanol

Question 3

aciditity of CA

Answer 3

C=O weakens O-H
the R-C=O - O (O-H breaks and H leaves)
carboxylate ion is stabilised because of the delocalised electrons, spreads negative charge and reduces charge density.
Less likely for the H+ to bond again, less attracted. more disassociated

Question 4

the effect of electron withdrawing groups on CA

Answer 4

furhter weaken O-H bond and extend the delocalistion of electrons
carboxylate ion can stablilised when O-H bond breaks and less attracted to H+ so won’t reform acid, more disassociated

Question 5

which one is more stronger acid; trichloroethanoic acid or ethanoic acid?

Answer 5

trichloroethanoic acid, Cl is an electron-withdrawing groups
3 Cl electronegative atoms
ethanoic acid is the weakest, CH3 is electron donating group, strengthens O-H group = donates negative charge towards the COO- group, concentrated negative and with H+

Question 6

can methanoic acid be oxidised??

Answer 6

yes, it is a stronger reducing agent than other carboxylic acids so it can undergo oxidation.

Question 7

oxidation of methanoic acid

Answer 7

HCOOH
warmed with mild oxidising agents like Tollens and Fehlings
Fehlings : Cu2+ becomes a Cu+ and produces a red ppt
Tollens : Ag+ to Ag silver metal
also works with stronger oxidising agents e.g KMnO4 (decolourise purple solution) and K2Cr2O7 (orange –> green)
HCOOH + [O] –> CO2 + H2O
HCOOH –> CO2 + 2e- + 2H+(aq)

Question 8

oxidation of ethanedioic acid

Answer 8

sulfuric acid with KMnO4; produces CO2 and H2O
solid ethanedioic acid is weighed to produce an accurate solution of a known concentration. tirtrated with KMnO4 in the titration. end point is determined when the addition of the purple KMnO4 solution into pink, all oxidisied
2MnO4 - + 6H+ + 5H2C2O4 –> 2Mn2+ + 10CO2 + 8H2O
Mn2+ acts as a catalyst, autocataylst (products catalyse reaction)

Question 9

make acyl chlorides

Answer 9

1. phosphorous chloride (PCl5)
2. phosphorous chloride (PCl3) heat
3. sulfur dichloride oxide (SoCl2) liquid product

Question 10

phosphorous (V) chloride reaction to ethanoic acid

Answer 10

CH3COOH + PCl5 –> CH3COCl + HCl + POCl3

HCl = white fumes

Question 11

phosphorous (III) chloride reaction to ethanoic acid

Answer 11

3CH3COOH + PCl3 –> 3CH3COCl + H3PO3

heat is required

Question 12

sulfur dichloride oxide to ethanoic acid

Answer 12

CH3COOH + SOCl2 –> CH3COCl + HCl + SO2
HCl = white fumes
CH3COCl is the only liquid product

Question 13

reactivity acyl chrlorides vs CA

Answer 13

acyl shloricdes
the electrons of the O and Cl draw it away from the C atom and gives it a large partial positive charge
open for nucleophiles attack
gives off white fumes of HCl

Question 14

acyl chlorides with water

Answer 14

mechanism:
water attaches to the c
c=o gives electron to become C-O-
C-Cl breaks and C=O forms again
COOH and HCl

Question 15

acyl chlorides with alchol

Answer 15

esters and HCl
acyl chlorides react faster than CA and go into completion (stronger acid)
C=O-Cl completely swapped with alcohol without the H
H and Cl forms HCl

Question 16

acyl chrlodies with ammonia

Answer 16

NH3 acts as nucleophile on acyl chlorides
the Cl is replaced with an NH2
forms amide and HCl
HCl + NH3 –> NH4+Cl-

Question 17

aryl chlorides hydrolysis

Answer 17

doesn’t happen
the lone pair of electrons of Cl overlaps with the delocalised p electrons in the benzene ring.
C-Cl double bond character which makes the bond stronger and harder to break

Question 18

acyl chlorides with phenol

Answer 18

phenol can’t react with CA so acyl chlorides are warmed for phenyl esters
phenol and base = base accepts H so forms O-
this phenoxide ion acts as nucleophile to attack the acyl chloride

Question 19

acyl chrlodies with amines

Answer 19

primary amines act as nucleophils as the nitrogen has a lone pair of electrons and attacks the acyl chloride
amine removes an H and replaces the Cl to form an amide and HCl

similarly in secondary amines.
both HCl formed are not white fumes. HCl acts as a base with unreacted amine to form an ammonium salt
CH3NH2 + HCl –> CH3NH3+Cl-
(CH3)2 NH2 + HCl –> (CH3)2NH3+Cl-

Question 20

forming a cooh

Answer 20

CH3CH2CN + HCl + 2H2O = CH3CH2COOH + NH4Cl

Question 21

reduction of COOH

Answer 21

CH3CH2COOH + 4[H] –> CH3CH2OH + H2O