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Politics

Russia says alleged US spy declared missing

September 24, 2019

Moscow said ex-Kremlin official Oleg Smolenkov, alleged to be a CIA spy, has officially been declared missing. Smolenkov reportedly had access to Vladimir Putin and was extracted in 2017 to avoid media exposure.

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Russian flag outside the building of the Embassy of the United States of America in Moscow
Image: Getty Images/AFP/V. Maximov

Kremlin employee Oleg Smolenkov has been declared missing by Moscow and is now wanted by the Russian authorities, the country's Interior Ministry said on Monday.

The person may have disappeared with his wife and three children while on holiday in Montenegro two years ago, according to reports in Russian daily Kommersant.

Previously, US broadcaster CNN and The New York Times newspaper reported on a high-level covert source inside the Kremlin, adding that said the spy was extracted from Russia in 2017 over concerns for his safety. They did not name the CIA informant.

Unconfirmed reports in Russia later named Smolenkov as a CIA spy.

Responding to those reports, Kremlin spokesman Dmitry Peskov said that Smolenkov did not have direct access to President Vladimir Putin and that he had been a low-level official "fired several years ago." 

Read more: Russia confirms alleged US spy worked in Kremlin

The spokesman said he had no information about Smolenkov working for the CIA. Peskov also downplayed the reports that the man had been extracted as "a kind of 'Pulp Fiction.'"

Earlier this month, Russia asked the US to clarify Smolenkov's whereabouts after Russian news outlets reported that a man with his name owned a house in the city of Stafford, in the US state of Virginia. Many former US military and ex-FBI employees live in the area, which is near Washington DC, according to Russia's RIA news agency.

kw,dj/rc (AP, Reuters, Interfax)

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