Question

Pre-Lab for Lidocaine Analog Synthesis Step 1 1) Complete the reaction table below with the expected quantities used and obtained in this reaction NH2 HN 1) ACOH 2) Na Chloroacetyl chloride 2,6- Sodium a-chloro-2,6- Acetic acid dimethylace tanilide dimethylaniline acetate 0.4 M (aq) FW g/mol g/mol g/mol g/mL g/mL Amount g (Theory) mL mL mL mL (Theory) mmol Equivs 1.0 1.1 2) Draw the mechan is m of this amide formation (you can look up nucleophilic acyl substitution reactio ns in textbook or Wikipedia)

Answer 1

1.

a 1. ACOH 2. NaOac . NaDAE CI CI CHUN Mol WL: 121.18 CH-0 Mol. Wt.: 112.94 CHOCNO Mol. Wt.: 197.66

	2,6-dimethylaniline	Chloroacetyl chloride	Acetic acid	Sodium acetate	α-chloro-2-6-dimethylacetanilide
FW	121.18 g/mole	112.94 g/mole	---	0.4 M (aq)	197.66 g/mole
d	0.984 g/ml	1.42 g/ml	---	---	---
Amount	1.0 ml	0.71 ml	5.0 ml	22.3 ml	1.60g (Theory)
mmole	8.12	8.932	---	8.932	8.12 (Theory)
Equivs.	1.0	1.1	---	1.1	---

Sodium Acetate = FW. 82.03 g/mole

1M NaOAc = 82.03g in 1000 ml H₂O

Therefore 4M NaOAc = 328.12g in 1000ml H₂O

Therefore 0.4M NaOAc = 32.812g in 1000ml H₂O

We have taken 1.1equivalent NaOAc for reaction which is 0.732g.

Therefore 22.3ml 0.4M solution contains 0.732g.

Here the calculation done on the basis of 1.0ml batch size of 2,6-dimethylamine.

Quantity of acetic acid taken as a solvent.

Calculation carried out by using these formule,

Mmole= (Amount(g)/FW) x 1000

Theoretical yield= {FW(Product)/FW(substrate)} x amount of substrate(g)

Ml quantity is converted to gram quantity using density.

2. Mechanism:-

NH 3 T HCI COACO