Propose a synthesis of triphenylmethanol, (c6h5)3coh, beginning with methyl benzoate, c6h5cooch3, and bromobenzene, c6h5br, via the intermediacy of an organometallic reagent.

Bromobenzene should be reacted as follows:

1. Mg(s) in THF

2. Methyl benzoate

3. H3O+

Explanation:

First, bromobenzene reacts with magnesium in the solvent THF to form the organometallic intermediate phenylmagnesium bromide, which is a strong nucleophile. Two equivalents of reacts with methyl benzoate by nucleophilic attack at the carbonyl carbon to form triphenylmethoxide, which is then protonated in the final step to form the desired product triphenylmethanol.

the correct answer is b. a strong acid completely dissociates to its ions when dissolved in water. completely dissociates to and ions. a is incorrect because does not fully ionize to and ions and some remain as the compound . c and d are incorrect because, and do not fall into the category of bronsted lowry acids. a bronsted lowry acid is a proton donor and since both molecules do not have a proton to donate they are not bronsted lowry acids. likewise,   if we use the lewis acid definition, they cannot be acids. a lewis acid is an electron pair acceptor. both and have a complete octet so can not accept any electrons which disqualifies them as acids.

answer : (a) volume of the gas = 8.67 ×   l

                (b) second virial coefficient b = -1.21226 l/mole

solution :   given,     t = 300k

                                p = 20atm

                                z = 0.86

                                n = 8.2 mmoles = 8.2 × moles

                                r =   0.082 l atm/k/mol

        find,     (a) volume of gas   and (b) second virial coefficient b at 300k.

formula used :            

                          and            

(a)            

                    = \times 0.082\times 300}{20}[/tex]

                    = 8.67 ×   l

(b)      

                    = (0.00867-1) × 8.659

                    = - 8.5839 l/mole