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PCAT Topic-wise Practice Test Questions - 2022 PCAT Comprehensive Writing Review Mock Tests Objective-type Multiple Choice Questions (MCQs) PsychCorp, a brand of Harcourt Assessment Pharmacy College Admission Test Exam 2022 PCAT Examination Details Name of the Exam

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Practice free online PCAT sample mock test quiz: Download Pharmacy College Admission Test solved model exam PDF papers with previous years' questions and answers. Eliminate (D).PCAT Exam 2022 Practice Test Questions: Pharmacy College Admission Test Quiz (D) As indicated above, thermolysis of Compound X results in formation of 1,3-Dimethylcyclohezene via a syn rather than anti elimination mechanism. A reasonable mechanism that accounts for the course of thermolysis of Compound X is shown below. The all- cis isomer (Compound Y) does not possess a syn β-hydrogen and thus fails to afford an alkene upon thermolysis. (C) Thermolysis of Compound X proceeds via a cyclic transition state with concomitant syn eliminaton. The fact that thermolysis of Compound Y fails to afford an alkene is inconsistent with a mechanism that proceeds via anti elimination. (B) Recall that E2 eliminations proceed in a single step with concomitant stereospecific anti elimination. However, these products were not obtained upon thermolysis of Y. The intermediate carbocation derived from either Compound X or Y is expected to lose a proton to afford 1,3-dimethylcyclohexene along with O=C-(SH)SCH 3. (A) Recall that E1 eliminations proceed via formation of a carbocationic intermediate. Attractive intermolecular London dispersion forces are expected to be more important for CO 2 than for H 2 and thus reduce the volume occupied by CO 2 relative to that occupied by an ideal gas. Electrons in CO 2 π-orbitals are more polarizable than those in, e.g., H 2, which are tightly held in the 1s orbital. (D): “Polarizability” offers a measure of the ease with which electron clouds can be distorted. (C): The net result of intermolecular interactions between CO 2 molecules at STP is that they are attractive and thus reduce the volume occupied by CO 2 relative to that occupied by an ideal gas. (B): Intermolecular van der Waals forces result in attractive interactions, which reduce the volume occupied by CO 2 at STP relative to that occupied by an ideal gas. (A): Due to molecular symmetry the C=O dipoles in CO 2 cancel one another hence, CO 2 possesses no net dipole moment. (ii) The assumption that there are no attractive intermolecular forces between gas molecules (i.e., that molecular collisions are perfectly elastic) breaks down. The correct response to this question is (D).ĭeviation from ideal gas behavior results for two main reasons: (i) The assumption that molecules of a gas are point-sized and thus occupy negligible volume within their container becomes invalid. (D) Ozonolysis of H 2C=CH(CH 2) 3CH=CH 2 is expected to afford two products: O=CH(CH 2) 3CH=O + H 2C=O (2 equivalents). (C) Ozonolysis of 1-methylcyclohexene is expected to proceed with concomitant ring opening to afford only one product, i.e., H 3C-C(O)-(CH 2) 4CH=O. (B) Ozonolysis of (H 3C) 2C=CH-CH 2CH=CH 2 is expected to afford three products: (H 3C) 2C=O + O=CH-CH 2-CH=O + H 2C=O. (A) Ozonolysis of H 3C-CH=C(CH 3)-CH 2CH=CH 2 is expected to afford three products: H 3C-CH=O + H 3C-C(O)-CH 2CH=O + H 2C=O. Ozonolysis of internal alkynes followed by aqueous workup affords carboxylic acids. Oxidative workup of the reaction mixture obtained via ozonlysis of alkenes produces carboxylic acids, whereas the corresponding reductive workup affords aldehydes.

Recall that ozonolysis is employed to cleave C=C double bonds in alkenes and carbon to carbon triple bonds in alkynes. Given the four compounds (A)–(D), Compound X can either be a diene or a cyclic monoalkene.

Inspection of the molecular formula C nH 2n-2 indicates that Compound X contains two degrees of unsaturation.

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