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Russia officially declares alleged CIA spy missing

Oleg Smolenkov disappeared while on holiday in Montenegro in 2017, says interior ministry

Adam Forrest
Monday 23 September 2019 15:13 BST
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The house in Stafford, Virginia, reportedly used by alleged spy Oleg Smolenkov
The house in Stafford, Virginia, reportedly used by alleged spy Oleg Smolenkov (AFP/Getty Images)

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Russian authorities have declared a former Kremlin official – alleged to have become a CIA informant – a missing person and revealed they are looking for him.

An entry in the Interior Ministry’s database shows officials are actively searching for Oleg Smolenkov, the RIA news agency reported on Monday.

Mr Smolenkov disappeared with his wife Antonina and three children while on holiday in Montenegro in June 2017, according to the Russian daily newspaper Kommersant.

US media reports have said a CIA informant in the Russian government was extracted and brought to the United States in 2017.

The reports, confirmed by Reuters, citing two sources, have described the Russian informant as a high-level CIA source.

The Kremlin has confirmed that Mr Smolenkov worked in the Russian presidential administration, but said he was fired in the 2016/17 period, but claimed he was not a senior official.

Russia asked the US via Interpol earlier this month to clarify the whereabouts of Mr Smolenkov after Russian news reports identified a house listed as belonging to a man with his name in the Stafford area of Virginia near Washington DC.

The area where the house is located is inhabited by many former US military and FBI personnel, according to the RIA.

Russia reportedly fired officials who allowed Mr Smolenkov to flee the country via Montenegro.

He trip violated a ban on employees of Vladimir Putin’s presidential administration travelling to Montenegro – introduced in the spring of 2017 after the Balkan country accused Moscow of trying to stage a coup.

Additional reporting by Reuters

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