The unknown in the first line is 5 grams of 3-isopropoxyaniline (C 9 H 13 NO) th
ID: 507368 • Letter: T
Question
The unknown in the first line is 5 grams of 3-isopropoxyaniline (C9H13NO) the molecular weight is (151.10). Have to find the product (xg). Need to see all calculations and all the unknowns that go in the highlighted areas. Please show all working, such as grams, moles, molecular weights (anything in the highlighted areas needed). would be great if you wrote back the question with all areas filled in (just copy and paste the transcribed text)
Experimental Using the mass provided for your unknown, calculate the highlighted quantities and rewrite the procedure below for your unknown W Nitration of unknown: In a 100 mL single neck RBF, concentrated sulfuric acid (15 mL), nitric acid (15 mL) and the unknown (x g, y mmol) were combined. The reaction was refluxed for 2 hours, and cooled to room temperature. The strongly acidic solution was careful poured onto ice-cold water (300 mL), neutralized with NaoH 1M) and extracted with diethyl ether (3 x 100 mL). The ether extracts were dried using magnesium sulfate. After filtering and removal of the volatile solvent, the crude mixture was subjected to column chromatography, ultimately yielding the desire product (x g, 59.1%)Explanation / Answer
Answer:
Nitration of 3-isopropoxyaniline: In a 100 mL single neck RBF, concentrated sulfuric acid (15mL), nitric acid (15 mL) and the 3-isopropoxyaniline (5 g, 33.10 mmol) were combined. The reaction was refluxed for 2 hours, and cooled to room temperature. The strongly acidic solution was careful poured onto ice-cold (300 mL), neutralized with NaOH (1 M) and extracted with diethyl ether (3 x 100 mL). The ether extracts were dried using magnesium sulfate. After filtering and removal of the volatile solvent, the crude mixture was subjected to column chromatography, ultimately yielding the 5-nitro-3-isopropoxyaniline (2.955 g, 59.1%)
There are certain points to note here:
(a.) In the first step, an aniline is taken in highly acidic condition. Therefore, due to +I effect of the ammonium ion, an incoming (-NO2)+ will be directed to an electron rich meta positions (i.e. either 3 or 5 positions). Since 3rd position is already occupied with iso-propyl group, therefore, instead of more general o- and p- product, a m- product formation at position 5' will predominate. This result in formation of 5-nitro-3-isopropoxyaniline as major product (59.1%).
(b.) After refluxing for 2 hrs in acidic medium a nitration at 2', 4' and 5' positions will occur. Note: Please see diethyle ether is mentioned as (3 x 100 mL) which indicate possiblity of nitration at all three positions. However, due to the reason mentioned in point (a) a nitration at position 5' will predominate. Consequently, after column chromatagraphy, 5-nitro-3-isopropoxyaniline will be formed in 59.1 %
(3.) On netralization with NaOH, an ammonium ion will convert or restore to the aniline moiety.
(4.) The volume of sulfuric acid, nitric acid, ice-cold water, NaOH and diethyl ether does not play any role in yield calculation, as they are just solvents which are taken in excess.
Calculation of y:
As Number of moles = Mass of the compund taken / Molar weight of the compound taken
=> number of moles of 3-isopropoxyaniline (y) = 5 g / 151.10 g = 0.033 mol
which can be converted to mmol by multiplying with 1000 i.e. 0.033 x 1000 = 33.10 mmol
Calculation of desired product (x g):
As the yield of the product formed (%) = [mass of the product formed / mass of the reactant taken ) ] x 100
Therefore:
59.1 % = (mass of 3-isopropoxyaniline / mass of 5-nitro-3-isopropoxyaniline) x 100
=> 59.1 % = (5 g / mass of 5-nitro-3-isopropoxyaniline) x 100
=> mass of 5-nitro-3-isopropoxyaniline (x g) = [(59.1 x 100) x 5] = 2.955 g
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